Finding the indices of duplicate values in one array

Given one array A=[ 1 1 2 3 5 6 7].

I need help to known the indices where there are duplicate values.

Thanks

Answer 1

[Stephan Koehler]

A = [1 2 3 2 5 3]
[v, w] = unique( A, 'stable' );
duplicate_indices = setdiff( 1:numel(A), w )

this should work too, and is elegant

[Jun W]

How about finding how many times are those elements repeated?

[Image Analyst]

Use histcounts and look for bins with more than 2 counts.

A = [1 2 3 2 5 3]
[counts, edges] = histcounts(A)

A =

1 2 3 2 5 3

counts =

1 2 2 0 1

edges =

Columns 1 through 5

0.5 1.5 2.5 3.5 4.5

Column 6

5.5

You can see that the bins for 2 and 3 both have 2 counts so there are multiples of 2 and 3 in A.

Note: This will find any repeats, and they don't have to be consecutive. If you want to look for consecutive repeats, call the diff() function and look for zeros.

Answer 2

[Image Analyst]

Here's one way:

A = [-2 0 1 1 2 3 5 6 6 6 7 11 40]
% Elements 3, 4, 8, 9, and 10 are repeats.
% Assume A is integers and get edges
edges = min(A) : max(A)
[counts, values] = histcounts(A, edges)
repeatedElements = values(counts >= 2)
% Assume they're integers
% Print them out and collect indexes of repeated elements into an array.
indexes = [];
for k = 1 : length(repeatedElements)
  indexes = [indexes, find(A == repeatedElements(k))];
end
indexes % Report to the command window.

You get [3,4,8,9,10] as you should.

[Image Analyst]

Arthur, with your new array A = [29892, 29051, 29051], my code still works. It returns 2 and 3. Can you tell me why you're still trying to use Adam's code even after I told you it doesn't work but mine does?

A = [29892, 29051, 29051]; 
% Elements 2 and 3 are repeats.
% Assume A is integers and get edges
edges = min(A) : max(A);
[counts, values] = histcounts(A, edges);
repeatedElements = values(counts >= 2)
% Assume they're integers
% Print them out and collect indexes of repeated elements into a cell array.
indexes = [];
for k = 1 : length(repeatedElements)
  indexes = [indexes, find(A == repeatedElements(k))];
end
indexes % Report to the command window.

[Arthur Souza]

because I don't have 'histcounts' function. Where do I find it?

[Image Analyst]

Then you have a version older than R2014b. What version do you have? This should work in old versions:

A = [29892, 29051, 29051]; 
% Elements 2 and 3 are repeats.
% Assume A is integers and get edges
edges = min(A) : max(A);
[counts, values] = histc(A, edges);
repeatedElements = edges(counts >= 2)
% Assume they're integers
% Print them out and collect indexes of repeated elements into an array.
indexes = [];
for k = 1 : length(repeatedElements)
  indexes = [indexes, find(A == repeatedElements(k))];
end
indexes % Report to the command window.

[Arthur Souza]

I have the 2013a version. and... IT WORKED! Thank you so much Image Analyst! I think my problem is solved now! Have a nice weekend!

[Tyann Hardyn]

You save my life (indirectly) again, Mr Image Analyst. Thank you so much. You helped someone else, then your help will be a good answer for the others, like me, lol.

Answer 3

[Adam]

[~, uniqueIdx] = unique( A );
duplicateLocations = ismember( A, find( A( setdiff( 1:numel(A), uniqueIdx ) ) ) );

then

find( duplicateLocations )

will give you the indices if you want them rather than a logical vector.

There are probably neater methods though.

If you want only the duplicates after the first then simply

setdiff( 1:numel(A), uniqueIdx )

should do the job.

[Arthur Souza]

I tried this:

[U,I]=unique(A(:,1)); repeated=setdiff(1:size(A,1),I)

but MATLAB returns me this -> Error using unique Too many input arguments. Error in setdiff>setdiffR2012a (line 505) c = unique(c,order); Error in setdiff (line 84) [varargout{1:nlhs}] = setdiffR2012a(varargin{:});

Can somebody help me? I'm new here

[Image Analyst]

Adam, this doesn't work in general:

A=[ 1 1 2 3 5 6 6 7]
[~, uniqueIdx] = unique( A );
duplicateLocations = ismember( A, find( A( setdiff( 1:numel(A), uniqueIdx ) ) ) )
indexes = find(duplicateLocations)

It gets the wrong indexes for the repeated 6's:

duplicateLocations =
  1×8 logical array
   1   1   1   0   0   0   0   0
indexes =
     1     2     3

It should give [1,2,6,7].

Arthur, your code worked for me for the A given. If it doesn't work for you, give us your A.

[Arthur Souza]

My A is an arbitrary vector, like this one you used here. I would like to know why this error occurs and try to fix it. I have to find these indexes to use them on another vector. Don't know why, but the A you showed here didn't work for me =/

[Arthur Souza]

I am trying with an A like this: A = [29892, 29051, 29051]; But it still doesn't wokr for me.

[Jan]

"Doesn't work" is a weak description of the problem. Please post the error message or explain the difference between the results and your expectations.

[Arthur Souza]

This is the error message -> Error using unique Too many input arguments. Error in setdiff>setdiffR2012a (line 505) c = unique(c,order); Error in setdiff (line 84) [varargout{1:nlhs}] = setdiffR2012a(varargin{:});

[Arthur Souza]

My problem is the same as the topic of this forum: Finding the indices of duplicate values in one array. I use the same solution that has been put here, but only this error message is returned to me

[CompViscount]

Commenting here as it's led me to overall the best answer here, it just has a mistake. The "find" in the 2nd line changes the values into indices before passing to ismember, which just makes the output nonsense. I removed that. Using the same numbers as image analyst above:

A=[ 1 1 2 3 5 6 6 7]

A = 1×8

1 1 2 3 5 6 6 7

[~, uniqueIdx] = unique(A);
dupeIdx = ismember( A, A( setdiff( 1:numel(A), uniqueIdx ) ) );
dupes = A(dupeIdx)

dupes = 1×4

1 1 6 6

dupeLoc = find(dupeIdx)

dupeLoc = 1×4

1 2 6 7

[Gabor]

This works, thanks

Answer 4

[Jan]

function Ind = IndexOfMultiples(A)
T   = true(size(A));
off = false;
A   = A(:);
for iA = 1:numel(A)
  if T(iA)          % if not switched already
    d = (A(iA) == A);
    if sum(d) > 1   % More than 1 occurrence found
       T(d) = off;  % switch all occurrences
    end
  end
end
Ind = find(~T);
end

If the input has more than 45 elements, this is faster:

function T = isMultiple(A)
% T = isMultiple(A)
% INPUT:  A: Numerical or CHAR array of any dimensions.
% OUTPUT: T: TRUE if element occurs multiple times anywhere in the array.
%
% Tested: Matlab 2009a, 2015b(32/64), 2016b, 2018b, Win7/10
% Author: Jan, Heidelberg, (C) 2021
% License: CC BY-SA 3.0, see: creativecommons.org/licenses/by-sa/3.0/

T        = false(size(A));
[S, idx] = sort(A(:).');
m        = [false, diff(S) == 0];
if any(m)        % Any equal elements found:
   m(strfind(m, [false, true])) = true;
   T(idx) = m;   % Resort to original order
end
end

[GONZALEZ DE COSSIO ECHEVERRIA Francisco Jose]

Any way to speed up this code? Thanks

[Muhammad Imran]

Thanks Jan

Answer 5

[MRINAL BHAUMIK]

A=[ 1 1 2 3 5 6 7 6]
B = A'./A
B = B-diag(diag(B))
[pos1 pos2]=find(B==1)

o/p
pos1 =

     2
     1
     8
     6

Answer 6

[Anamika]

In MATLAB, you can find the indices of duplicate values in an array using the `find` function along with the `unique` function. Here's how you can do it:

A = [1 1 2 3 5 6 7];

% Finding the unique elements in the array

unique_elements = unique(A);

% Initializing an empty array to store the indices of duplicate values

duplicate_indices = [];

% Iterating through each unique element

for i = 1:numel(unique_elements)

% Finding the indices of occurrences of the current unique element

indices = find(A == unique_elements(i));

% If there are more than one occurrence, add the indices to the duplicate_indices array

if numel(indices) > 1

duplicate_indices = [duplicate_indices indices];

end

end

% Displaying the indices of duplicate values

disp(duplicate_indices);

Running this code will give you the indices of the duplicate values in the array A. In this case, the output will be: 1 2

This means that the duplicate values are located at indices 1 and 2 in the array A.

Answer 7

[Eduardo Gonzalez Rodriguez]

Here is my solution to find repeated values and their counts

function [dup, counts] = duplicates(A)
    [dup,~,n] = unique(A, 'rows', 'stable');
    counts = accumarray(n, 1, [], @sum);
    dup(counts==1) = [];
    counts(counts==1) = [];

Answer 8

[Piotr]

Hello,

here is my attempt to solve it. I faced similar problem but in my case I wanted to have the result in two column representation. Each row contains indices of repeated values.

A = [ 1 1 2 3 5 6 7 6];
nk = nchoosek(1:length(A),2);
nk(diff(A(nk),[],2)~=0,:) = [];
disp(nk)

Cheers, Piotr